substance.
Properties We take the density of water to be 1000 kg/m3.
Analysis The pressure difference between two points 1 and 2 in an incompressible fluid is given by P2−P1=−ρax(x2−x1)−ρ(g+az)(z2−z1) or P1−P2 =ρ(g+az)(z2−z1)
Therefore, tank A has a higher pressure at the bottom.
Discussion We can also solve this problem quickly by examining the relation Pbottom =ρ(g+az)h. Acceleration for tank B is about 1.5 times that of Tank A (14.81 vs 9.81 m/s2), but the fluid depth for tank A is 4 times that of tank B (8 m vs 2 m). Therefore, the tank with the larger acceleration-fluid height product (tank A in this case) will have a higher
ax
θ = 12°
Water tank
3-104
Solution A water tank is being towed on an uphill road at constant acceleration. The angle the free surface of water makes with the horizontal is to be determined, and the solution is to be repeated for the downhill motion case.
Assumptions 1 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 2 The acceleration remains constant.
Analysis We take the x- and z-axes as shown in the figure. From geometrical considerations, the horizontal and vertical components of acceleration are
α
The tangent of the angle the free surface makes with the horizontal is 4078
When the direction of motion is reversed, both ax and az are in negative x- and z-direction, respectively, and thus become negative quantities,
α
Then the tangent of the angle the free surface makes with the horizontal becomes
5801
Discussion Note that the analysis is valid for any fluid with constant density, not just water, since we used no information that pertains to water in the solution.
z
3-105
Solution A vertical cylindrical tank open to the atmosphere is rotated about the centerline. The angular velocity at which the bottom of the tank will first be exposed, and the maximum water height at this moment are to be determined.
Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 Water is an incompressible fluid.
Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as
( 2 )
) 4
( 0 2 R2 r2
h g r
zs = −ω −
where h0 = 0.3 m is the original height of the liquid before rotation. Just before dry spot appear at the center of bottom surface, the height of the liquid at the center equals zero, and thus zs(0) = 0. Solving the equation above for ω and substituting,
rad/s
≅7.62
=
=
= 7.625rad/s
m) 45 . 0 (
m) 3 . 0 )(
m/s 81 . 9 ( 4 ] 4
2 2 2
0
R ω gh
Noting that one complete revolution corresponds to 2π radians, the rotational speed of the container can also be expressed in terms of revolutions per minute (rpm) as
⎟=72.8rpm
⎠
⎜ ⎞
⎝
= ⎛
= 1min
s 60 rad/rev 2
rad/s 625 . 7
2π π
n& ω
Therefore, the rotational speed of this container should be limited to 72.8 rpm to avoid any dry spots at the bottom surface of the tank.
The maximum vertical height of the liquid occurs a the edges of the tank (r = R = 0.45 m), and it is
= + = + =0.600m
) m/s 81 . 9 ( 4
m) 45 . 0 ( rad/s) 625 . 7 ) ( m 3 . 0 4 ( )
( 2
2 2
2 2
0 g
h R R
zs ω
Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property.
ω
0.9 m z
0 r
3-106
Solution A cylindrical tank is being transported on a level road at constant acceleration. The allowable water height to avoid spill of water during acceleration is to be determined.
Assumptions 1 The road is horizontal during acceleration so that acceleration has no vertical component (az = 0).
2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant.
Analysis We take the x-axis to be the direction of motion, the z-axis to be the upward vertical direction, and the origin to be the midpoint of the tank bottom. The tangent of the angle the free surface makes with the horizontal is
4077 . 0 0 81 . 9
tan 4 =
= +
= +
z x
a g
θ a (and thus θ = 22.2°)
The maximum vertical rise of the free surface occurs at the back of the tank, and the vertical midplane experiences no rise or drop during acceleration. Then the maximum vertical rise at the back of the tank relative to the midplane is
cm 8.2 m 082 . 0 0.4077 m)/2]
40 . 0 [(
tan ) 2 /
max =( = × = =
Δz D θ
Therefore, the maximum initial water height in the tank to avoid spilling is cm
=51.8
−
= Δ
−
= tank max 60 8.2
max h z
h
Discussion Note that the analysis is valid for any fluid with constant density, not just water, since we used no information that pertains to water in the solution.
D=40 cm ax= 4 m/s2
θ
htank =60 cm Δz
Water tank
3-107
Solution A vertical cylindrical container partially filled with a liquid is rotated at constant speed. The drop in the liquid level at the center of the cylinder is to be determined.
Assumptions 1 The increase in the rotational speed is very slow so that the liquid in the container always acts as a rigid body. 2 The bottom surface of the container remains covered with liquid during rotation (no dry spots).
Analysis Taking the center of the bottom surface of the rotating vertical cylinder as the origin (r = 0, z = 0), the equation for the free surface of the liquid is given as
( 2 )
) 4
( 2 2
2
0 R r
h g r
zs = −ω −
where h0 = 0.6 m is the original height of the liquid before rotation, and rad/s
57 . s 12 60
min rev/min) 1 120
( 2
2 ⎟=
⎠
⎜ ⎞
⎝
= ⎛
= π π
ω n&
Then the vertical height of the liquid at the center of the container where r = 0 becomes
0.44m
) m/s 81 . 9 ( 4
m) 20 . 0 ( rad/s) 57 . 12 ) ( m 0 6 . 0 4 ( )
0
( 2
2 2
2 2
0− = − =
= g
h R
zs ω
Therefore, the drop in the liquid level at the center of the cylinder is Δhdrop,center =h0−zs(0)=0.60−0.44=0.16 m
Discussion Note that the analysis is valid for any liquid since the result is independent of density or any other fluid property. Also, our assumption of no dry spots is validated since z0(0) is positive.
z
r ω
zs
R = 20 cm Free
surface
ho = 60 cm
g
3-108
Solution The motion of a fish tank in the cabin of an elevator is considered. The pressure at the bottom of the tank when the elevator is stationary, moving up with a specified acceleration, and moving down with a specified acceleration is to be determined.